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drag the tiles to the correct boxes to complete the pairs. how many sol…

Question

drag the tiles to the correct boxes to complete the pairs.
how many solutions does each of these equations have?
one solution two solutions no solutions
|2x + 1| = 0
|x + 21| = 2
2|x - 2| = -1

Explanation:

for \(|2x + 1| = 0\):

Step1: Recall absolute value property

The absolute value equation \(|A| = 0\) implies \(A = 0\) (since the only number with absolute value 0 is 0 itself). So for \(|2x + 1| = 0\), we set \(2x+1 = 0\).

Step2: Solve for \(x\)

\(2x+1 = 0\)
Subtract 1 from both sides: \(2x= - 1\)
Divide by 2: \(x =-\frac{1}{2}\)
This gives one solution.

for \(|x + 21| = 2\):

Step1: Recall absolute value property

The absolute value equation \(|A| = B\) (where \(B>0\)) implies \(A = B\) or \(A=-B\). Here \(B = 2>0\), so for \(|x + 21| = 2\), we have two cases: \(x + 21=2\) or \(x + 21=-2\).

Step2: Solve each case

  • Case 1: \(x + 21=2\)

Subtract 21 from both sides: \(x=2 - 21=-19\)

  • Case 2: \(x + 21=-2\)

Subtract 21 from both sides: \(x=-2 - 21=-23\)
This gives two solutions.

for \(2|x - 2| = -1\):

Step1: Simplify the equation

Divide both sides by 2: \(|x - 2|=-\frac{1}{2}\)

Step2: Recall absolute value property

The absolute value of a number is always non - negative (i.e., \(|A|\geq0\) for any real number \(A\)). But here we have \(|x - 2|=-\frac{1}{2}\), and \(-\frac{1}{2}<0\). Since the absolute value can never be negative, there are no solutions.

Answer:

  • \(|2x + 1| = 0\) \(

ightarrow\) one solution

  • \(|x + 21| = 2\) \(

ightarrow\) two solutions

  • \(2|x - 2| = -1\) \(

ightarrow\) no solutions

So the pairings are:
\(|2x + 1| = 0\) \(
ightarrow\) one solution
\(|x + 21| = 2\) \(
ightarrow\) two solutions
\(2|x - 2| = -1\) \(
ightarrow\) no solutions